Question: $f\,^{\prime}(x)=-5e^x$ and $f(3)=22-5e^3$. $f(0) = $
Finding $f(x)$ We have $f'(x)=-5e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-5e^x)\,dx \\\\ & = {-5e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(3)=22-5e^3$. Here's what we get when we plug in $3$ : $\begin{aligned}f(3)&={-5e^{3}} {+ C} \end{aligned}$ We are given that this must equal $22-5e^3$ : $22-5e^3 = {-5e^{3}} {+ C}$ Solving the equation gives us ${C=22}$. Finding $f(0)$ Now, we have that $f(x)={-5e^x} {+ 22}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=-5e^0 + 22\\\\ &=17 \end{aligned}$ The answer $f(0) = 17$